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I am stuck on this one. I know since $(a_n)$ is Cauchy, then $ \forall \epsilon>0$ $\exists n,m \in \mathbb[N]$ so that whenever $n,m \geq N, |a_n - a_m| < \epsilon$.

So I'm thinking, when $n$ is odd, $|-a_n + a_m| < \epsilon$, and when

$n$ is even, $|a_n - a_m| < \epsilon$.

I'm not sure if I'm on the right track here or not.

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    $\begingroup$ No it is not Cauchy. $\endgroup$ – Mhenni Benghorbal Mar 11 '16 at 7:22
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    $\begingroup$ On intuition, the alternating sequence should converge to the same limit (if it has a limit), because every even n gives just the sequence itself. Similarly, every odd n give the negative sequence, both are converging subsequences. Therefore, when s is the limit, we should have -s = s, so s = 0, note that this is not rigorous. As an exercise you can show the following rigorously: When an is a nonnegative cauchy sequence and (-1)^n an is a cauchy sequence, then a converges to 0. $\endgroup$ – SometimesBlind Mar 11 '16 at 7:28
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The sequence $1,1,1,1,\dots$ is a Cauchy sequence; $1,-1,1,-1,\dots$ is not.

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  • $\begingroup$ For some reason I had completely forgotten about that. I was thinking about monotonic converging sequences! $\endgroup$ – Jabernet Mar 11 '16 at 20:44
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    $\begingroup$ Kind of a trick question, because the first Cauchy sequences one thinks of are the ones converging to $0$. And for those, the statement is true. $\endgroup$ – André Nicolas Mar 11 '16 at 22:53
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The correct version of the statement would read as:

Let $a_n$ be a cauchy sequence. Then, $c_n = (-1)^n a_n$ is a cauchy sequence if and only if $\lim_{n\rightarrow \infty} a_n = 0$.

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